12 th BM MATRIX 2 MARKS TEST 2 - WordPress.com
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12 th BM MATRIX 2 MARKS TEST 2 12th Standard 2019 EM Business Maths ***************************************** RAVI MATHS TUITION CENTER, GKM COLONY, CH-82. PH- 8056206308 Date : 09-Jun-19 Reg.No. : Time : 00:30:00 Hrs Total Marks : 20 10 x 2 = 20 Show that the equations5x+3y+7z=4,3x+26y+2z=9,7x+2y+10z =5 are consistent and solve them by rank method. The price of three commodities X,Y and Z are x,y and z respectively Mr.Anand purchases 6 units of Z and sells 2 units of X and 3 units of Y. Mr.Amar purchases a unit of Y and sells 3 units of X and 2units of Z. Mr.Amit purchases a unit of X and sells 3 units of Y and a unit of Z. In the process they earn `5,000/-, `2,000/- and `5,500/- respectively Find the prices per unit of three commodities by rank method. Solve the following equations by using Cramer’s rule 2x + 3y = 7; 3x + 5y = 9 A total of Rs 8,600 was invested in two accounts. One account earned 4 3 4 % annual interest and the other earned 6 1 2 % annual interest. If the total interest for one year was Rs 431.25, how much was invested in each account? (Use determinant method). A total of Rs 8,500 was invested in three interest earning accounts. The interest rates were 2%, 3% and 6% if the total simple interest for one year was Rs 380 and the amount ,invested at 6% was equal to the sum of the amounts in the other two accounts, then how much was invested in each account? (use Cramer’s rule). Two types of soaps A and B are in the market. Their present market shares are 15% for A and 85% for B. Of those who bought A the previous year, 65% continue to buy it again while 35% switch over to B. Of those who bought B the previous year, 55% buy it again and 45% switch over to A. Find their market shares aer one year and when is the equilibrium reached? Find the rank of the matrix A = −2 1 3 0 1 1 1 3 4 4 2 7 ( ) Find k if the equations 2x+3y−z=5,3x−y+4z=2,x+7y−6z=k are consistent. The cost of 2kg. of wheat and 1kg. of sugar is Rs 100. The cost of 1kg. of wheat and 1kg. of rice is Rs 80. The cost of 3kg. of wheat, 2kg. of sugar and 1kg of rice is Rs 220. Find the cost of each per kg., using Cramer’s rule. Solve the following equation by using Cramer’s rule x + 4y + 3z =2,2x−6y + 6z=−3, 5x− 2y + 3z =−5 10 x 2 = 20 Given non-homogeneous equations are 5x+ 3y + 7z = 4 3x + 26y + 2z = 9 7x + 2y + 10z = 5 The matrix equation corresponding to the given system is 5 3 7 3 26 2 7 2 10 x y z = 4 9 5 Augmented matrix [A, B] Elementary Transformation ( )( ) ( ) 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 1) www.Padasalai.Ne www.Padasalai.Ne www.Padasalai.Ne www.Padasalai.Ne www.Padasal www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasal www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasal www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasal www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasal www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasal www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasal www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasal www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasal www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasal www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasal www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasal www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasal ww.Padasalai.Net ww.Padasalai.Net ww.Padasalai.Net ww.Padasalai.Net ww.Padasal www.Padasalai.N www.Padasalai.N www.Padasalai.N www.Padasalai.N www.Padasala www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasala www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasala www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasala www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasala www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasala www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasala www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasala www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasala www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasala www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasala www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasala www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasalai.Net www.Padasala ww.Padasalai.Net ww.Padasalai.Net ww.Padasalai.Net ww.Padasalai.Net ww.Padasala Padasalai
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Transcript of 12 th BM MATRIX 2 MARKS TEST 2 - WordPress.com
12 th BM MATRIX 2 MARKS TEST 2
12th Standard 2019 EM
Business Maths
*****************************************
RAVI MATHS TUITION CENTER, GKM COLONY, CH-82. PH- 8056206308
Date : 09-Jun-19
Reg.No. :
Time : 00:30:00 Hrs Total Marks : 2010 x 2 = 20
Show that the equations5x+3y+7z=4,3x+26y+2z=9,7x+2y+10z =5 are consistent and solve them by rank method.
The price of three commodities X,Y and Z are x,y and z respectively Mr.Anandpurchases 6 units of Z and sells 2 units of X and 3 units of Y. Mr.Amar purchases aunit of Y and sells 3 units of X and 2units of Z. Mr.Amit purchases a unit of X andsells 3 units of Y and a unit of Z. In the process they earn `5,000/-, `2,000/- and`5,500/- respectively Find the prices per unit of three commodities by rank method.
Solve the following equations by using Cramer’s rule 2x + 3y = 7; 3x + 5y = 9
A total of Rs 8,600 was invested in two accounts. One account earned 434% annual
interest and the other earned 612
% annual interest. If the total interest for one year
was Rs 431.25, how much was invested in each account? (Use determinant method).
A total of Rs 8,500 was invested in three interest earning accounts. The interest rates were 2%, 3% and 6% if the total simpleinterest for one year was Rs 380 and the amount ,invested at 6% was equal to the sum of the amounts in the other twoaccounts, then how much was invested in each account? (use Cramer’s rule).
Two types of soaps A and B are in the market. Their present market shares are 15% for A and 85% for B. Of those whobought A the previous year, 65% continue to buy it again while 35% switch over to B. Of those who bought B the previousyear, 55% buy it again and 45% switch over to A. Find their market shares a�er one year and when is the equilibriumreached?
Find the rank of the matrix A =
−2 1 30 1 11 3 4
427( )
Find k if the equations 2x+3y−z=5,3x−y+4z=2,x+7y−6z=k are consistent.
The cost of 2kg. of wheat and 1kg. of sugar is Rs 100. The cost of 1kg. of wheat and1kg. of rice is Rs 80. The cost of 3kg. of wheat, 2kg. of sugar and 1kg of rice is Rs 220.Find the cost of each per kg., using Cramer’s rule.
Solve the following equation by using Cramer’s rulex + 4y + 3z =2,2x−6y + 6z=−3, 5x− 2y + 3z =−5
10 x 2 = 20
Given non-homogeneous equations are
5x+ 3y + 7z = 4
3x + 26y + 2z = 9
7x + 2y + 10z = 5
The matrix equation corresponding to the given system is
5 3 73 26 27 2 10
xyz
=
495Augmented matrix [A, B] Elementary Transformation
( ) ( ) ( )
1)
2)
3)
4)
5)
6)
7)
8)
9)
10)
1)
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Padasalai
Augmented matrix [A, B] Elementary Transformation
5 3 73 26 27 2 10
495
∼
3 26 25 3 77 2 10
945
R1 ↔ R2
1263
23
5 3 77 2 10
345
R1 → R1 ÷ 3
1263
23
0−1213
113
7 2 5
3−115
R2 → R2 − 5R1
1263
23
0−1213
113
0−1763
163
3−11−16
R3 → R3 − 7R1
∼
1263
23
0−113
13
0−113
13
3−1−1
R2 → R2 ÷ 11
R3 → R3 ÷ 16
1263
23
0−113
13
0 0 0
3−10
R3 → R3 − R2
Here ρ(A) = ρ(A, B) = 2 < Numberofunknowns.
∴ The system is consistent with infinitely many solutions let us rewrite the above echelon form into matrix form
1263
23
0−113
13
0 0 0
xyz
=
3−10
x +26
3y +
2
3z = 3
26 2
( )( )
( )( )( )( )
( )( )( ) ( )
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Padasalai
x +26
3y +
2
3z = 3
|letz = k where kE R
(2) ⇒−11
3y +
k
3= − 1
⇒ − 11y = − 3 − k 11y=3+k
⇒ y =1
11(3 + k)
Substituting y =
1
11(3 + k)
and z = kin (1) we get,
x +26
3
3 + k
11+
2
3k = 3
=26
3
3 + k
11−
2k
3+ 3
78 − 26k
33−
2k
3+ 3
78 − 26k − 22k + 99
33
78 − 26k − 22k + 99
33
21 − 48k
33=
3(7 − 16k)
33
= 1
11(7 − 6k)
∴
Solution set is 1
11(7 − 16k).
1
11(3 + k), k
K ϵ
R
Hence, for di�erent values of k, we get infinitely many solutions.
( )( )
{ }Given that the price of commodities X, Y and Z are x, y and z respectively
By the given data
Transaction x y z Earning
Mr. Anand +2+3-6 Rs.5000
Mr. Amar +3-1 +2Rs.2000
Mr. Amit -1 +3+1Rs.5500
Here, purchasing is taken as negative symbol and selling is taken as positive symbol
Thus, the non-homogeneous equations are
2x + 3y - 6z = 5000
3x - y + 2z = 2000
-x + 3y + z = 550
The matrix equation corresponding to the given system is
2 3 −63 −1 2−1 3 1
xyz
=
500020005500
Augmented matrix Elementary TransformationAugmented matrix Elementary Transformation
( ) ( ) ( )
2)
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Padasalai
2 3 −63 −1 2−1 3 1
500020005500
−1 3 13 −1 22 3 −6
550020005000
1 −3 −13 −1 20002 3 −6
−500020005500
R1 → R1( − 1)
1 −3 −10 8 50 9 −4
−55001850016000
R2 → r2 − 3R1
R3 → R3 − 2R1
1 −3 −1
0 1658
0 1−49
−5500185008
160009
R2 → R2 ÷ 8
R3 → R3 ÷ 9
1 −3 −1
0 158
0 0−49 −
58
−5500185008
160009 −
18500
8
R3 → R3 − R2
1 −3 −1
0 158
0 0−7772
−5500185008
−3850072
.Clearly the last equivalent matrix is in echelon form and it has three non-zero rows . . :. ∴ ρ(A) = ρ([A, B]) = 3 Number of
unknowns. ∴ The given system is consistent and has unique solution. To find the solution, let us rewrite the above : echelon
form into the matrix form.
1 −3 −1
0 158
0 0−7772
xyz
−5000185008
−3850072
⇒ x − 3y − z = − 5500
y +5
8z =
18500
8
−77
72z =
38500
72
( )( )( )( )
( )( )( )
( )( ) ( )
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Padasalai
⇒ z =−38500
−77
⇒ z = 500
(2) ⇒ y +5
8(500) =
18500
8 y =
18500
8−
2500
8
⇒ y = 2000
(1) ⇒ x − 3(2000) − 500 = − 5500
⇒ x − 6000 − 500 = − 5500
⇒ x − 6000 − 500 = − 5500
⇒ x = − 5500 + 6500
⇒ x = 10000
Hence, the prices per unit of three commodities are Rs1000, Rs 2000 and Rs500 respectively
Δ =2 33 5 = 10 − 9 = 1 ≠ 0
Since Δ ≠ 0 we can apply Cramer's rule and the system is consistent with unique solution.
Δx =7 39 5 = 7(5) − 9(3)
= 35 - 27 = 8
Δy =2 73 9 = 2(9) − 3(7)
= 18 - 21 =-3
∴
x =
Δx
Δ=
8
1= 8
y =Δy
Δ=
−3
1= 3
∴ Solution set is {8, -3)
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Padasalai
Let the amount invested in the two accounts be Rs x and Rs. y respectively
By the given data, x + y = 8600 ..(1)
43
4×
x
100+ 6
1
2×
y
100= 431.25
∴ interest =
PNR
100
⇒19x
400+
13y
3200= 431.25
⇒19x + 26y
400= 431.25
19x + 26y = 172500 ...(2)
Δ =1 119 26 = 1(26) − 1(19)
= 26-19 =7
Δx =8600 1
172500 26 = 8600(26) − 1(172500)
= 223600 - 172500= 51100
Δy =1 860019 172500 = 1(172500) − 19(8600)
= 172500 - 163400 = 9100
x =Δx
Δ−
51100
7= 7300
y =Δy
Δ=
9100
7= 1300
∴ Investment in the interest of
43
4
% account is Rs. 7300 and investment in the rate of 61
2
account is Rs. 1300.
[ ]
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Let the amount invested in the rate of2%, 3% and 6% be Rs. x, Rs. y and Rs. z respectivly
By the given data,
x+ y+z = 8500
2x
100+
3y
100+
6z
100= 380
⇒2x + 3y + 6z
100= 380
∵ Interest =PNR
100=
x × 1 × 2
100=
2x
100
⇒ 2x + 3y + 6z = 38000
Also,z = x+y
x+y-z =0
Δ =
1 1 12 3 61 1 −1
13 61 −1 − 1
2 61 −1 + 1
2 31 1
= 1(-3 - 6) - 1(-2 -6) + 1 (2 - 3)
= 1(-9) - 1 (-8) + 1(-1)
= -9+8-1=2 ≠ 0
Since Δ ≠ 0 Cramer's rule can be applied and the system is consistent with unique solution
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Padasalai
Since Δ ≠ 0,Cramer s rule can be applied and the system is consistent with unique solution
Δx =
8500 1 138000 3 6
0 1 −1
= 8500
3 61 −1 − 1
38000 61 −1 + 1
38000 30 1
= 8500 (- 3 - 6) - 1(-38000 -0) + 1(38000 - 0)
= 8500(-9) - 1(-38000) + 1(38000)
= - 76500 + 38000 + 38000
= -500
Δy =
1 8500 12 38000 61 0 −1
= 1
38000 60 −1 − 8500
2 61 −1 + 1
2 380001 0
= 1 (-38000 - 0) - 8500 (-2 -6) + 1(0 - 38000)
= - 38000 - 8500 (-8) - 38000
= - 38000 + 68000 - 38000
= - 8000
Δz =
1 1 85002 3 380001 1 0
= 1
38000 60 −1 − 1
2 380001 0 + 8500
2 31 1
= 1 (0 - 38000) - 1(0 -38000) +85000 (2 - 3)
= - 38000 + 38000 + 8500 (-1)
= - 8500
Hence, the amount invested in the three accounts are Rs. 250, Rs. 4000 and Rs. 4250 respectively.
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Padasalai
Transition probability matrix
(A B) T = (A B)
Where A represents the percent of people those who bought soap A and B represents the percent of people those who bought
soap B.
By the given data
A = 15%=·15
and B = 85% = ·85
Percentage a�er one year is
( ⋅ 15 ⋅ 85)⋅ 65 ⋅ 35⋅ 45 ⋅ 55
= ((.15)(·65) + (·85)(-45) ·15(-35)+ ·85(-55))
= (-0975 + ·3825 ·0525 + -4675)
= (-48 ·52)
Hence, market share a�er one year is 48% and 52%
At equilibrium,
(A B)⋅ 65 ⋅ 35⋅ 45 ⋅ 55 = (A B)
(-65A+A5B ·35A+·55B) = (A B)
Equating the corresponding entries on both sides
we get
⇒ ⋅ 65A + ⋅ 45B = A
⇒ ⋅ 65A + ⋅ 45(1 − A) = A
[Since A+B = 1 =} B= 1-A]
⇒ ⋅ 65A + ⋅ 45 − ⋅ 45A = A
⇒ ⋅ 45 = A − ⋅ 65A + ⋅ 45A⇒ ⋅ 45 = A( ⋅ 35 + 45)⇒ ⋅ 45 = A( ⋅ 35 + 45)⇒ ⋅ 45 = A( − 8)
⇒ A =⋅ 45
⋅ 8= ⋅ 5625 = 56.25
∴ B = 1 − A = 1 − ⋅ 5625 = ⋅ 4375= 43.75%
∴ Equilibrium is reached when A = 56.25% and B = 43.75%
( )
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Padasalai
Given
A =−2 1 30 1 11 3 4
427
∼1 3 40 1 1−2 1 3
724
R1 → R3
∼1 3 40 1 10 7 11
7218
R2 → R2 + 2R1
∼1 3 40 1 10 0 4
724
R3 → R3 − 7R2
The last equivalent matrix is in echelon form and there are 3 non - zero rows.
∴ ρ(A) = 3
( )( )( )( )
Given non-homogeneous equations are
2x + 3y - z = 5, 3x - y + 4z = 2, x + 7y - 6z = k
Augmented matrix Elementary Transformation
2 3 −13 −1 41 7 −6
52k
1 7 −63 −1 42 3 −1
k25
R1 ↔ R3
1 7 −60 −22 220 −11 11
k2 − 3k5 − 2k
R2 → R2 − 3R1
1 7 −60 −22 220 0 0
k2 − 3k
2(5 − 2k) − (2 − 3k)
R2 → R2 − 3R1
R3 → R3 − 2R1
1 7 −60 −22 220 0 0
k2 − 3k
10 − 4k − 2 + 3k
R3 → 2R3 − R2
−1 7 −60 −22 220 0 0
k2 − 3k8 − k
Here ρ(A) = 2
Since the given system is consistent, ρ(A, B) must
be equal to 2.
This can happen only when
8-k=0 ⇒ k=8
( )( )( )( )( )( )
7)
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Padasalai
Let the cost of lkg of wheat be Rs. x, lkg of sugar be Rs. y and lkg of rice be Rs. z.
By the given data,
2x+ Y = 100
x +z = 80
3x+ 2y+z = 220
Δ =
2 1 01 0 13 2 1
= 20 12 1 − 1
1 13 1 + 0
= 2 (0 - 2) -1 (1 - 3) + 0
= 2(-2) - 1(- 2)
= - 4 + 2 =-2
Δx =
100 1 080 0 1220 2 1
= 1000 12 1 − 1
80 1220 1 + 0
= 100(0 - 2) - 1 (80 - 220)
= 100(- 2) - 1(- 140)
= - 200 + 140 = - 60.
Δy =
2 100 01 80 13 220 1
= 280 1220 1 − 100
1 13 1 + 0
= 2 (80 - 220) - 100 (1 - 3)
= 2 (- 140) - 100 (-2)
= - 280 + 200 = - 80.
Δz =
2 1 1001 0 803 2 220
20 802 220 − 1
1 803 220 + 100
1 03 2
= 2(0 - 160) - 1(220 - 240) + 100(2 - 0)
= 2(- 160) - 1(- 20) + 100(2)
= - 320 + 20 + 200
= -100
x =Δx
Δ=
−60
−2= 30
y =Δy
Δ=
−80
−2= 40
z =Δz
Δ=
−100
−2= 50
∴ The cost of lkg of wheat is Rs.30
The cost of lkg sugar is Rs.40 and
The cost of 1 kg of rice is Rs.50
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Padasalai
Δ =
1 4 32 −6 65 −2 3
= 1
−6 6−2 3 − 4
2 65 3 + 3
2 −65 −2
= 1(-18 + 12) - 4(6 - 30) +3 (- 4 +30)
= 1(- 6) - 4(- 24) + 3(26)
= - 6 + 96 + 78 = 168 ≠ 0
Since Δ ≠ 0 the system is consistent with unique solution and Cramer's rule can be applied
Δx =
2 4 3−3 −6 6−5 −2 3
= 2
−6 6−2 3 − 4
−3 6−5 3 + 3
−3 −6−5 −2
= 2 (- 18 + 12) - 4(- 9 +30) + 3(6 -30)
= 2(- 6) - 4(21) + 3(- 24)
= -12-84-72=-168
Δy =
1 2 32 − 65 −5 3
= 1−3 6−5 3 − 2
2 65 3 + 3
2 −35 −5
= 1 (-9+30)-2(6-30)+3(- 10+ 15)
= 1(21) - 2(- 24) + 3(5)
= 21 + 48 + 15 = 84
Δz =
1 4 22 −6 −35 −2 −5
= 1
−6 −3−2 −5 − 4
2 −35 −5 + 2
2 −65 −2
= 1(30-6)-4(-10+ 15)+2(-4+30)
= 24 - 4(5) + 2(26)
= 24 - 20 + 52 = 56
Solution set is − 1,
12 ,
1 ,3
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Padasalai
